Solution:

Combining the two fractions, the numerator is $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$ and the denominator is $9 \cdot 11 \cdot 3 \cdot 5 \cdot 7.$ These are the same product.

So the value is $1.$

Thus, the correct answer is A .

Solution:

There are $10$ terms. The average of the first and last is $\dfrac{90 + 99}{2} = 94.5.$

So the sum is $10 \cdot 94.5 = 945.$

Thus, the correct answer is B .

Solution:

Since $\dfrac{10^7}{10^4} = 10^3 = 1000,$ the expression is $\dfrac{1000}{5} = 200.$

Thus, the correct answer is D .

Solution:

Completing the figure to the full $6 \times 9$ rectangle gives an area of $54.$ The piece removed from the lower-left corner is a rectangle measuring $2$ by $4,$ with area $8.$

So the polygon's area is $54 - 8 = 46.$

Thus, the correct answer is C .

Solution:

The satisfactory grades number $5 + 4 + 3 + 3 = 15,$ and the total number of grades is $15 + 5 = 20.$

So the fraction that is satisfactory is $\dfrac{15}{20} = \dfrac{3}{4}.$

Thus, the correct answer is C .

Solution:

Since $7.5$ cm is $1.5$ times $5$ cm, the stack holds $1.5$ times as many sheets.

That is $1.5 \times 500 = 750$ sheets.

Thus, the correct answer is D .

Solution:

Row $n$ contains $2n - 1$ squares. Because both ends are white and the colors alternate, each row has one more white square than black square, so the number of black squares is $n - 1.$

For the $37$th row, this is $37 - 1 = 36.$

Thus, the correct answer is C .

Solution:

Substituting $a = -2$ gives the values $-3a = 6,$ $4a = -8,$ $\dfrac{24}{a} = -12,$ $a^2 = 4,$ and $1.$

The largest of these is $6,$ which is $-3a.$

Thus, the correct answer is A .

Solution:

Each factor $1 - \dfrac1k$ equals $\dfrac{k-1}{k},$ so the product is

$$\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10}.$$

Every numerator cancels the previous denominator, leaving $\dfrac{1}{10}.$

Thus, the correct answer is A .

Solution:

The midpoint of $\dfrac15$ and $\dfrac13$ is their average.

$$\dfrac{\frac15 + \frac13}{2} = \dfrac{\frac{8}{15}}{2} = \dfrac{4}{15}.$$

Thus, the correct answer is C .

Solution:

Fold the net so that $X$ is the bottom face. Then $U,$ $V,$ $W,$ and $Z$ wrap around to become the four side faces.

The only remaining square, $Y,$ becomes the top face, which is opposite the bottom face $X.$

Thus, the correct answer is E .

Solution:

The triangle's perimeter is $6.2 + 8.3 + 9.5 = 24$ cm, so the square also has perimeter $24$ cm.

The square's side is $\dfrac{24}{4} = 6$ cm, so its area is $6^2 = 36 \text{ cm}^2.$

Thus, the correct answer is B .

Solution:

Walking: $4 \times \dfrac34 = 3$ miles. Running: $10 \times \dfrac12 = 5$ miles.

The total distance is $3 + 5 = 8$ miles.

Thus, the correct answer is B .

Solution:

The two taxes differ by $0.5\%$ of the price.

That is $0.005 \times \$20 = \$0.10.$

Thus, the correct answer is B .

Solution:

All $100$ numbers from $200$ to $299$ contain a $2.$

Among $100$ to $199,$ there are $10$ with a $2$ in the tens place and $10$ with a $2$ in the units place, but $122$ is counted twice, giving $10 + 10 - 1 = 19.$ The range $300$ to $399$ similarly contributes $19.$

The total is $100 + 19 + 19 = 138.$

Thus, the correct answer is C .

Solution:

The $30$ students split into $5$ equal parts of $6.$ Boys make up $2$ parts ($12$) and girls $3$ parts ($18$).

So there are $18 - 12 = 6$ more girls than boys.

Thus, the correct answer is D .

Solution:

Seven tests averaging $85$ total $7 \times 85 = 595$ points; six tests averaging $84$ total $6 \times 84 = 504$ points.

The seventh score is $595 - 504 = 91.$

Thus, the correct answer is D .

Solution:

From $9p \lt 10$ we get $p \lt 1.111\ldots,$ and from $10p \gt 11$ we get $p \gt 1.10.$

The only price in whole cents between $\$1.10$ and $\$1.111\ldots$ is $\$1.11.$

Thus, the correct answer is E .

Solution:

The new perimeter is $2(1.1\ell + 1.1w) = 1.1 \cdot 2(\ell + w),$ which is $1.1$ times the old perimeter.

That is a $10\%$ increase.

Thus, the correct answer is B .

Solution:

Since $31 = 4 \cdot 7 + 3,$ the weekdays falling on January $1, 2,$ and $3$ each occur five times that month, and every other weekday occurs four times.

For Tuesday and Saturday to occur only four times, neither may be January $1, 2,$ or $3.$ The only starting day that works is Wednesday: then Wednesday, Thursday, and Friday occur five times, while Tuesday and Saturday each occur four times.

Thus, the correct answer is C .

Solution:

After four raises the salary is multiplied by $1.1^4 = 1.4641.$

That is an increase of $46.41\%,$ which is more than $45\%.$

Thus, the correct answer is E .

Solution:

The first digit is one of $8$ allowed digits ($2$ through $9$), so $\dfrac18$ of the numbers begin with $9.$ The last digit is any of $10$ digits, so $\dfrac{1}{10}$ end in $0.$

These conditions are independent, so the fraction is $\dfrac18 \cdot \dfrac{1}{10} = \dfrac{1}{80}.$

Thus, the correct answer is B .

Solution:

Each day there are $1200 \times 5 = 6000$ student-class attendances. Since each class holds $30$ students, there are $\dfrac{6000}{30} = 200$ classes.

Each teacher teaches $4$ classes, so there are $\dfrac{200}{4} = 50$ teachers.

Thus, the correct answer is E .

Solution:

Adding the three side sums gives $3S = (10 + 11 + \cdots + 15) + (\text{sum of corners}) = 75 + (\text{sum of corners}).$ To maximize $S,$ place the three largest numbers $13, 14, 15$ at the corners, giving corner sum $42.$

Then $3S = 75 + 42 = 117,$ so $S = 39.$ This is achievable: with corners $13, 14, 15$ and midpoints $12, 10, 11,$ each side sums to $39.$

Thus, the correct answer is D .

Solution:

To show the rule "a vowel forces an even number" is false, Mary needs a card with a vowel on one side and an odd number on the other. No card shows a vowel, since $P$ and $Q$ are consonants.

The only card showing an odd number is $3.$ Turning it over reveals a vowel, which contradicts Jane's claim.

Thus, the correct answer is A .